[二分]codeforces 274A k-Multiple Free Set

k-Multiple Free Set

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A k-multiple free set is a set of integers where there is no pair of integers where one is equal to another integer multiplied by k. That is, there are no two integers x and y (x < y) from the set, such that y = x·k.

You're given a set of n distinct positive integers. Your task is to find the size of it's largest k-multiple free subset.

Input

The first line of the input contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109). The next line contains a list of n distinct positive integers a1, a2, ..., an (1 ≤ ai ≤ 109).

All the numbers in the lines are separated by single spaces.

Output

On the only line of the output print the size of the largest k-multiple free subset of {a1, a2, ..., an}.

Examples

input

Copy

6 2
2 3 6 5 4 10

output

Copy

3

Note

In the sample input one of the possible maximum 2-multiple free subsets is {4, 5, 6}.

 题意:给你n个数，找出在这n个数中最多有多少个x使得所选出的数中比x大的数不是x的k倍

注意：先选择所有的数，若x的k倍在原序列存在，则看x的k倍的k倍是否存在，若存在则删去x的k倍，因为这样就可以选x和x的k倍的k倍，若x的k倍的k倍不存在，则删x的k倍

排序后二分查找，k和a[i]最大是1e9，所以要注意k*k*a[i]是否小于1e18，为防止溢出可用if(k*a[i]<=(1e18)/k)

自己的测试数据：

2 1000000000
1000000000 1000000000

4 2
2 3 4 8

5 1
1 2 3 4 5

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int amn=1e5+5;
const ll inf=1e18;

int n,k,mid;
ll a[amn];

ll fd(ll b){
    int l=1,r=n;
    mid=(l+r)/2;
    while(l<r){
        if(a[mid]<b)l=mid+1;
        else if(a[mid]>b) r=mid-1;
        else break;
        mid=(l+r)/2;
    }
    return a[mid];
}

int main(){
    ios::sync_with_stdio(0);
    cin>>n>>k;
    for(int i=1;i<=n;i++){
        cin>>a[i];
    }
    sort(a+1,a+1+n);
    int ans=n,pos,jg,jg1;
    if(k!=1)
    for(int i=1;i<n;i++){
        if(!a[i])continue;
        jg=fd(k*a[i]);
        if(k*a[i]==jg){
            pos=mid;
            if(k*a[i]<=inf/k){
                jg1=fd(k*k*a[i]);
                if(k*k*a[i]==jg1){
                a[pos]=0;
                }
                else a[i]=0;
            }
            else a[i]=0;
            ans--;
        }
    }
    printf("%d
",ans);
}